Problem 4 a) Sei \(a d-b c \neq 0\) voraus... [FREE SOLUTION] (2024)

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Chapter 2: Problem 4

a) Sei \(a d-b c \neq 0\) vorausgesetzt. Falls \(c=0\), ist \(d \neq 0\), also \(cx+d \neq 0\). Falls \(c \neq 0\), ist ebenfalls \(c x+d \neq 0\), denn andernfallswäre \(x=-c^{-1} d\) rational im Widerspruch zur Voraussetzung. Die Zahl $$ y:=\frac{a x+b}{c x+d} $$ ist also wohldefiniert. Mit $$ u:=a x+b, \quad v:=c x+d \neq 0 $$ erh?lt man. $$ u=v y, \quad d u-d v=(a d-b c) x, \quad-c u+a v=a d-b c \neq 0 $$ also $$ x=\frac{d u-b v}{-c u+a v}=\frac{d y-b}{-c y+a} $$ Wäre \(y\) rational, so auch \(x, \mathrm{im}\) Widerspruch zur Voraussetzung.Also ist \(y\) irrational.b) Sei \(a d-b c=0 .\) Wir können voraussetzen, dass \(cx+d \neq 0\) gilt und haben zu zeigen, dass \(y=(a x+b)(c x+d)^{-1}\) rationalist. Ist \(c \neq 0\), so folgt \(b=\frac{a d}{c}\), also $$ y=\frac{a x+b}{c x+d}=\frac{a x+\frac{a d}{c}}{c x+d}=\frac{a}{c} \in Q $$ Ist \(c=0\), so folgt \(d \neq 0\) und \(a=0\), also $$ y=\frac{b}{d} \in \mathbb{Q} $$

Short Answer

Expert verified

Under the condition that \(a d-b c \neq 0\), the number \(y=\frac{a x+b}{c x+d}\) is well-defined and \(y\) is irrational. However, if assumed that \(a d-b c=0\), then \(y\) is rational.

Step by step solution

01

Break down the problem

Start by following the initial assumption that \(a d-b c \neq 0\). Assume further that if \(c=0\), then \(d \neq 0\) so \(c x+d \neq 0\), and once again if \(c \neq 0\), then \(c x+d \neq 0\). Now the important part is to set the 'well defined' number \(y=\frac{a x+b}{c x+d}\)

02

Demonstrate the well-definiteness of the number

Take \(u:=a x+b\), \(v:=c x+d\). You get \(u=v y\), \(d u-d v=(a d-b c) x\), and \(-c u+a v=a d-b c\), which gives \(x=\frac{d u-b v}{-c u+a v}=\frac{d y-b}{-c y+a}\). Thus, if \(y\) is rational, then \(x\) would be as well which contradicts the assumption. Hence the number \(y\) is irrational in this condition.

03

Test a different condition

Start with the condition that \(a d-b c=0\). Current task is to show that \(y=(a x+b)(c x+d)^{-1}\) is rational. If \(c \neq 0\), then \(b=\frac{a d}{c}\), which gives \(y=\frac{a x+b}{c x+d}=\frac{a x+\frac{a d}{c}}{c x+d}=\frac{a}{c}\), that is a rational number. If \(c=0\), this gives \(d \neq 0\) and \(a=0\), and so, \(y=\frac{b}{d}\) and this also results in a rational number.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proof of Number Properties

Understanding how to prove properties of numbers, such as rationality or irrationality, is a fundamental skill in mathematics. The proof typically takes the route of logically establishing the truth of a proposition. In our exercise, we prove whether a certain number, represented as a quotient involving variables, is rational or irrational.

Initially, the proof hinges on the assumption that certain product differences are non-zero. Further assumptions about the numerator and denominator being non-zero ensure that the constructed fraction is valid. For mathematical proofs, such preliminary conditions help to prevent undefined operations and potential contradictions further down the argument.

The proof then continues by deducing properties of an equation derived from these assumptions. If a contradiction arises—like determining that a number must be both rational and irrational—then one of the initial assumptions must be false, leading to new insights about the system in question.

Rational Numbers

A rational number is defined as a number that can be expressed as the quotient or fraction of two integers, with the denominator not equal to zero. This set of numbers is represented by the symbol \( \mathbb{Q} \). When dealing with rational numbers, we look for this particular quotient structure.

In the given exercise, we examined a quotient involving the variables \( x \) to determine its rationality. We were able to show that under specific conditions, this quotient simplifies to the form that meets the definition of a rational number. This simplification process is crucial for declaring whether an expression, which at first glance might not resemble a fraction of integers, actually falls within the realm of rational numbers.

Mathematical Induction

While our original problem did not explicitly involve mathematical induction, it is worth understanding as a powerful proof technique often used to establish the truth of infinitely many cases. It involves proving a base case and then proving that if one case holds, the next logical case holds as well.

Though not used here, the principle of induction can be especially useful when dealing with sequences or sets of numbers whose properties are tied to their position within a particular order. It is a cornerstone of proof strategies in mathematics and can be used to demonstrate properties of both rational and irrational numbers in broader contexts.

Analysis Mathematics

Analysis mathematics, or simply 'analysis,' refers to the rigorous study of numbers and functions. It is concerned with limits, continuity, and the behavior of sequences. In the context of our exercise, analysis could be applied to investigate the limits of sequences involving \( y \) as \( x \) approaches certain values.

Moreover, in analysis, we often deal with proving the irrationality or rationality of numbers. This is closely related to the concept of algebraic numbers (solutions to polynomial equations with rational coefficients) and transcendental numbers (numbers not algebraic). The exercise demonstrates a method to determine whether numbers derived from variable expressions are algebraic or transcendental through logical reasoning and simplification - a process that is at the heart of mathematical analysis.

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Problem 4 a) Sei \(a d-b c \neq 0\) voraus... [FREE SOLUTION] (3)

Most popular questions from this chapter

\( \mathbf{H} .\) Es sei \(M:=N \cup\\{\infty\\}\), wobei \(\infty \notin\mathbb{N}\), Auf \(M\) führen wir zwei Verknüpfungen $$ \left\\{\begin{array} { l } { M \times M \longrightarrow M } \\ { ( a , b ) \longmapsto a + b } \end{array} \quad \text { und } \quad \left\\{\begin{array}{c} M \times M \longrightarrow M \\ (a, b) \longmapsto a \cdot b \end{array}\right.\right. $$ wie folgt ein: (1) Für \(a, b \in \mathbb{N}\) sei \(a+b\) bzw. \(a \cdot b\) die übliche Additionbzw. Multiplikation natürlicher Zahlen. (2) Für \(a \in M\) sei \(a+\infty=\infty+a=\infty\), (3) Flir \(a \in M \backslash\\{0\\}\) sei \(a \cdot \infty=\infty \cdota=\infty\). (4) \(0 \cdot \infty=\infty \cdot 0=0\) Man zeige, dass diese Verknüpfungen auf \(M\) die Körperaxiome \((\mathrm{A}.1),(\mathrm{A} .2)\), (A.3), (M.1), (M.2), (M.3) und (D), aber nicht (A.4) und(M.4) erfüllen.a) Nach Definition (vgl. An. \(1,(2.9))\) ist die Behauptung $$ \frac{a}{b}=\frac{c}{d} $$ gleichbedeutend mit $$ b^{-1} a=d^{-1} c $$ Multipliziert man beide Seiten der Gleichung (1) mit \(b d\), erhält man daraus $$ \underbrace{(b d)\left(b^{-1} a\right)}_{-: L}=\underbrace{(b d)\left(d^{-1}c\right)}_{-: R} $$ Durch wiederholte Anwendung der Axiome der Multiplikation (II.I) bis (II.4)ergibt sich \(L=(b d)\left(b^{-1} a\right) \stackrel{(\mathrm{II} .2)}{=}(d b)\left(b^{-1}a\right) \stackrel{(\mathrm{II}, 1)}{=} d\left(b\left(b^{-1} a\right)\right)\stackrel{(\mathrm{II}, 1)}{=} d\left(\left(b b^{-1}\right) a\right)\) \(\stackrel{(\mathrm{II} .4)}{=} d(1 \cdot a)\stackrel{\left(\mathrm{1l}_{1}\right)}{=} d a \stackrel{(\mathrm{II} .2)}{=}\mathrm{ad} .\)
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Problem 4 a) Sei \(a d-b c \neq 0\) voraus... [FREE SOLUTION] (2024)
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